Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was thinking of.

Searching for Mean(ing)

Stage: 3 Challenge Level:

Most of you sent us the right answer to the first question of the problem ('How many of 3kg and 8kg weights would you need for the average (mean) of the weights to be 6kg?'). Some found it by trial and error, for example, Ellen from Shincliffe Primary:

It is all about trial and error. I thought I had the answer when I had fourof 3kg weights and threeof 8kg weights. But then I noticed that I had seven weights. Then I kept my three of 8kg which is 24kg and all I had to do was add 2 of 3kg weights. And I had my answer.
3 x 8kg + 2 x 3kg =30, and 30/5 = 6kg.

Andy (Garden International School) pointed out:

The weight averages are from 3 to 8

Rosie from St Bartholomew's Cof E Primary School gave us the answers to some whole-number averages in between 3 and 8:

For an average of 7, you would need 4 of 8kg weights and 1 of 3kg weight.
For an average of 6, you would need 9 of 8kg weights and 6 of 3kg weights.
For an average of 5, you would need 2 of 8kg weights and 3 of 3kg weights.

Mr Judge's year 10 class, from School 21, made a spreadsheet to try lots of different combinations of the weights and find the possible values. You can see their spreadsheet here. What patterns can you see in their results? Can you explain them?

A general solution was provided by Hyeon (British School Muscat):

Imagine that the lighter weight is a and the heavier weight is b.
As long as a < b, the smallest average you can get is a and the biggest average you can get is b.
It is possible to get every single number in between. There are [(b-a)+1] averages.

Hyeon also noticed something important from the results of her trial with different weights:

If the total number of averages is odd, then 1 of a and 1 of b would give the middle average weight.

The amount of a and the amount of b used should add up to a factor or a multiple of the difference between a and b for the average to be a whole number.

Here are some of her trials:

For the 1kg and 5 kg weights

1 kg

5 kg

Total

Average

1

0

1

1

3

1

8

2

1

1

6

3

1

3

16

4

0

1

5

5

For the 17 kg and 57 kg weights

17 kg

57 kg

Total

Average

9

1

210

21

13

27

1760

44

1

7

1020

52

Although we received many answers to this problem, most of the answers simply stated the results rather than providing a general strategy for finding them. However, Anurag, from Queen Elizabeth's Grammar School in Horncastle, did draw some general conclusions. You can see his workinghere

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